The given expression is \( (x + y)^{100} + (x - y)^{100} \). To find the total number of terms in the expansion of this expression, we will first analyze the individual expansions of \( (x + y)^{100} \) and \( (x - y)^{100} \), and then combine them.
The binomial expansion of \( (x + y)^{100} \) is given by:
\[ (x + y)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^{100-k} y^k \]
This expansion has \( 101 \) terms because the index \( k \) ranges from 0 to 100, inclusive. So, the expansion of \( (x + y)^{100} \) contains 101 terms.
The binomial expansion of \( (x - y)^{100} \) is given by:
\[ (x - y)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^{100-k} (-y)^k = \sum_{k=0}^{100} \binom{100}{k} (-1)^k x^{100-k} y^k \]
Like the first expansion, this also has \( 101 \) terms. However, the signs of the terms alternate depending on whether \( k \) is even or odd due to the factor \( (-1)^k \). So, the expansion of \( (x - y)^{100} \) also contains 101 terms.
Now, consider the sum of the two expansions:
\[ (x + y)^{100} + (x - y)^{100} \] When we add these two expansions, terms where the powers of \( y \) are odd will cancel each other out, because \( (-y)^k \) will have the opposite sign of \( y^k \) when \( k \) is odd. On the other hand, terms where \( k \) is even will add up, because both expansions will have the same sign for those terms.
Thus, only the terms where \( k \) is even will remain after addition. The even values of \( k \) range from 0 to 100, inclusive. These values are \( k = 0, 2, 4, \dots, 100 \), which gives a total of 51 even values.
The total number of terms in the expansion of \( (x + y)^{100} + (x - y)^{100} \) is 51, because only the terms with even values of \( k \) survive after adding the two expansions
\[ \left( \frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1} \right) \left( \frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2} \right) \cdots \left( \frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}} \right) = \frac{\alpha^{13}}{{}^{14}C_0 \, {}^{14}C_1 \cdots {}^{14}C_{12}} \]
Then \[ 30\alpha = \underline{\hspace{1cm}} \]
200 ml of an aqueous solution contains 3.6 g of Glucose and 1.2 g of Urea maintained at a temperature equal to 27$^{\circ}$C. What is the Osmotic pressure of the solution in atmosphere units?
Given Data R = 0.082 L atm K$^{-1}$ mol$^{-1}$
Molecular Formula: Glucose = C$_6$H$_{12}$O$_6$, Urea = NH$_2$CONH$_2$